Fourier transform for biologist
A biologist’s way to learn Fourier transform
Visual intuition in 3D
This is an awesome introduction
Fourier Series
Discrete Fourier transform (DFT)
A Fourier series is a periodic function composed of harmonically related sinusoids, combined by a weighted summation.
周期性函数可以变换为正余弦函数的和
$$ \begin{aligned} f(t) &= \frac {a_0}{2} + \sum_n a_n \sin(n\omega t + \varphi_n) \cr &= \frac {a_0}{2} + \sum_n a_n \sin(n\omega t) + \sum_n b_n \cos(n\omega t) \end{aligned} $$
Note: 3 orthogonal bases ( 1, sin, cos )
Fourier Transform
Euler’s formula
For any real number $x$,
$$ e^{i\varphi} = \cos \varphi + i \sin \varphi $$
$i$ is the imagenary unit. let $\varphi = \omega t$, get complex exponentials
$$ e^{i\omega t} = \cos (\omega t) + i \sin (\omega t) $$
clockwise roation: $e^{i\omega t}$
counter-clockwise roation: $e^{ - i\omega t}$
when $\varphi = \pi$, Eluer's identity
$$ e^{i\pi} + 1 = 0 $$
Fourier transform for non-periodic function
For non-periodic function $f(t)$, we multiply $e^{ - i\omega t}$ to get signals only present in $e^{ - i\omega t}$, then
$$ \int_{-\infty}^{+\infty} f(t) e^{ - j\omega t} dt \rightarrow \begin{cases} = 0, \text{without } \omega \cr \neq 0, \text{with } \omega \end{cases} $$
Note: the bases of $e^{ - j\omega t}$ are orthogonal, if a signal is orthogonal to these bases, their dot product equal to 0.
- if a signal in $f(t)$ not present in $e^{ - j\omega t}$: $$\int_{-\infty}^{+\infty} f(t) e^{ - j\omega t} dt = 0$$
- if a signal in $f(t)$ present in $e^{ - j\omega t}$: $$\int_{-\infty}^{+\infty} f(t) e^{ - j\omega t} dt \neq 0$$
Now, the Fourier transform definition:
$$ F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{ - j\omega t} dt $$
and inverse Fourier transform:
$$ f(t) = \int_{-\infty}^{+\infty} F(\omega) e^{ - j\omega t}d \omega $$
2D Fourier transform
$$ F(u, v) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x,y) e^{ - j(ux + vy)} dxdy $$
Now, the orthogonal bases become (1, $\sin(ux + vy)$, $\cos(ux + vy)$), which could be further decomposited into:
- $\sin(ux)\sin(vy)$
- $\sin(ux)\cos(vy)$
- $\cos(ux)\sin(vy)$
- $\cos(ux)\cos(vy)$
Common application
- audio: $t$ is the time domain
- image: now $t$ become the location in the image
- low freq: contour
- high freq: detail
图像频率特性分析:
频谱图上的每一个像素点都代表一个频率值,幅值由像素点亮度变码而得。对于一幅图像,图像信号的频率特性如下:
- 直流分量: 表示预想的平均灰度
- 低频分量: 代表了大面积背景区域和缓慢变化部分
- 高频分量: 代表了它的边缘、细节、跳跃部分以及颗粒噪声
- 振幅: 描述了图像灰度的亮度
- 相位: 决定了图像是什么样子
Laplace Transform
Euler’s Formula:
$$ e^{i\omega t} = \cos (\omega t) + i \sin (\omega t) $$
we get
$$ \cos(\omega t) = \frac{1}{2} (e^{j\omega t} + e^{-j\omega t}) $$
$$ j\sin(\omega t) = \frac{1}{2} (e^{j\omega t} - e^{-j\omega t}) $$
Fourier transform:
$$ F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{ - j\omega t} dt $$
The problem of Fourier transformation is that each component of sinusoids keep constant magnitude while oscillating.
For $f(t)$ like $y = x^2$, when $x \rightarrow \infty$, FT do not perform well
To solve this problem, we could multiply a decay fuction $e^{-\sigma t}$, $\sigma > 0$.
$$ \begin{aligned} F(\omega) &= \int_{-\infty}^{+\infty} f(t) e^{-\sigma t} e^{ - j\omega t} dt \cr &= \int_{-\infty}^{+\infty} f(t) e^{-t(\sigma + j\omega)} dt \end{aligned} $$
With a decay fuction, FT perform well when $x \rightarrow \infty$.
Let complex number $s = \sigma + j\omega$, then Laplace Transform
is
$$ F(s) = \int_{-\infty}^{+\infty} f(t) e^{-st} dt $$
Laplace Transform:
- A generalized Fourier transform
- The magnitude keep increasing/decreasing as oscillation continue.
- if the real componet of $s$ is 0 ( $\sigma = 0$ ), then the magnitude will stay constant
Inverse Laplace transform:
$$ f(t) = \frac {1}{2 \pi i} \int_{c - i\infty}^{c + i\infty} F(s)e^{st} ds $$
A visual intution of Laplace Transform:
Fourier Transform and Convolution
Convolution: $$ h(x) = (f \star g)(x) = \int_{-\infty}^{+\infty} f(u)(g(x-u))du $$
discrete form:
$$ h[n] = (f \star g)[n] = \sum_{u = -\infty}^{+\infty} f[u]g[n-u] $$
时域卷积定理:时域上的卷积对应频域上的乘积
$$ F[f(t) \star g(t)] = F_f(\omega) \cdot F_g (\omega) $$
频域卷积定理:频域内的卷积对应时域内的乘积
$$ F[f(t) \cdot g(t)] = \frac {1} {2\pi} F_f(\omega) \star F_g (\omega) $$
Examlple:
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the output is
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Summary:
Weighted sum => convolution => multiplication after fourier tranform.
A short animation explained what convolution is